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Math Help - Positive integer proof

  1. #1
    Newbie darknight's Avatar
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    Positive integer proof

    If  n is a natural number, then
     (2003 + \frac{1}{2})^n + (2004 + \frac{1}{2})^n is a positive integer:

    A) when n is even
    B) when n is odd
    C) only when n=117 or n=119
    D) only when n=1 or n=3

    ====================================

    Simplifying the expression,
    (\frac{4007^n + 4009^n}{2^n})
    Therefore, (4007^n + 4009^n) must be divisible by (2^n)
    But how do I proceed from here? Can it be done without the binomial theorem?
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  2. #2
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by darknight View Post
    If  n is a natural number, then
     (2003 + \frac{1}{2})^n + (2004 + \frac{1}{2})^n is a positive integer:

    A) when n is even
    B) when n is odd
    C) only when n=117 or n=119
    D) only when n=1 or n=3

    ====================================

    Simplifying the expression,
    (\frac{4007^n + 4009^n}{2^n})
    Therefore, (4007^n + 4009^n) must be divisible by (2^n)
    But how do I proceed from here? Can it be done without the binomial theorem?
    This might not be the best solution, but we can use the multiple choice nature to our advantage.

    n = 1 is easily verified true, which rules out A and C. In order to test between B and D, all we need is to check n = 5.

    We want to know what is (4007^5 + 4009^5) modulo 2^5. So, reduce to (7^5 + 9^5) which turns out to be congruent to 16 mod 32. (On paper we can do 7 * 49 * 49 + 9 * 81 * 81 which becomes 7 * 17 * 17 + 9 * 17 * 17 which is just 16 * 17 * 17 which is clearly divisible by 16 but not 32.) So the answer is D.
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  3. #3
    Senior Member MacstersUndead's Avatar
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    Could you explain how you simplified the expression? I'm a bit rusty, but I tried the binomial theorem, and 'simplified' the equation to get

    \sum\binom{n} {k}[(2003)^{n-k} + (2004)^{n-k}](1/2)^k
    with k = 0 to n.

    how the question sets up the answers, the next step might be to seperate this sum into two sums, one with odd k terms and the other with even k terms.

    again, I'm rusty. blame summer haha.
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  4. #4
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by MacstersUndead View Post
    Could you explain how you simplified the expression? I'm a bit rusty, but I tried the binomial theorem, and 'simplified' the equation to get

    \sum\binom{n} {k}[(2003)^{n-k} + (2004)^{n-k}](1/2)^k
    with k = 0 to n.

    how the question sets up the answers, the next step might be to seperate this sum into two sums, one with odd k terms and the other with even k terms.

    again, I'm rusty. blame summer haha.
    Actually the binomial theorem was not used for that simplification; write 2003 + 1/2 as 4007/2, likewise with 2004 + 1/2, then distribute the exponent to numerators and denominators, then the two terms have a common denominator of 2^n
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  5. #5
    Senior Member eumyang's Avatar
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    Quote Originally Posted by darknight View Post
    If  n is a natural number, then
     (2003 + \frac{1}{2})^n + (2004 + \frac{1}{2})^n is a positive integer:

    A) when n is even
    B) when n is odd
    C) only when n=117 or n=119
    D) only when n=1 or n=3

    ====================================

    Simplifying the expression,
    (\frac{4007^n + 4009^n}{2^n})
    Therefore, (4007^n + 4009^n) must be divisible by (2^n)
    But how do I proceed from here? Can it be done without the binomial theorem?
    Note that:
    4007^1 ends with a 7,
    4007^2 ends with a 9,
    4007^3 ends with a 3,
    4007^4 ends with a 1,
    4007^5 ends with a 7,
    etc., etc.

    ... and:
    4009^1 ends with a 9,
    4009^2 ends with a 1,
    4009^3 ends with a 9,
    etc., etc.


    EDIT: Oops, a little late there...
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  6. #6
    Newbie darknight's Avatar
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    Thanks guys!
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